Week homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign ersions of these problems, arious details hae been changed, so that the answers will come out differentl. The method to find the solution is the same, but ou will need to repeat part of the calculation to find out what our answer should hae been. WebAssign Problem 1: In a football game a kicker attempts a field goal. The ball remains in contact with the kicker s foot for.5 s, during which time it eperiences an acceleration of 34 m/s. The ball is launched at an angle of 51 aboe the ground. Determine the horizontal and ertical components of the launch elocit. REASONING To determine the horizontal and ertical components of the launch elocit, we will use trigonometr. To do so, howeer, we need to know both the launch angle and the magnitude of the launch elocit. The launch angle is gien. The magnitude of the launch elocit can be determined from the gien acceleration and the definition of acceleration gien in Equation 3.. a = SOLUTION According to Equation 3., we hae t t or 34 m / s = m / s.5 s c hb g or = 34 m / s.5 s Using trigonometr, we find the components to be c hb g c hb g = cos 51 = 34 m / s.5 s cos 51 = 11 m / s = sin 51 = 34 m / s.5 s sin 51 = 13 m / s WebAssign Problem : Michael Jordan, formerl of the Chicago Bulls basketball team, had some fanatic fans. The claimed that he was able to jump and remain in the air for two full seconds from launch to landing. Ealuate this claim b calculating the maimum height that such a jump would attain. For comparison, Jordan s maimum jump height has been estimated at about one meter. REASONING AND SOLUTION The maimum ertical displacement attained b a projectile is gien b Equation 3.6b ( = + a ) with = : = a
In order to use Equation 3.6b, we must first estimate his initial speed. When Jordan has reached his maimum ertical displacement, =, and t = 1. s. Therefore, according to Equation 3.3b ( = +), a t with upward taken as positie, we find that = at = ( 9.8 m/s ) (1. s) = 9.8 m/s Therefore, Jordan's maimum jump height is (9.8 m/s) = = 4.9 m ( 9.8 m/s ) This result far eceeds Jordan s maimum jump height, so the claim that he can remain in the air for two full seconds is false. WebAssign Problem 3: A major-league pitcher can throw a baseball in ecess of 41. m/s. If a ball is thrown horizontall at this speed, how much will it drop b the time it reaches a catcher who is 17. m awa from the point of release? REASONING The ertical displacement of the ball depends on the time that it is in the air before being caught. These ariables depend on the -direction data, as indicated in the table, where the + direction is "up." -Direction Data a t? 9.8 m/s m/s? Since onl two ariables in the direction are known, we cannot determine at this point. Therefore, we eamine the data in the direction, where + is taken to be the direction from the pitcher to the catcher. -Direction Data a t +17. m m/s +41. m/s? Since this table contains three known ariables, the time t can be ealuated b using an equation of kinematics. Once the time is known, it can then be used with the -direction data, along with the appropriate equation of kinematics, to find the ertical displacement.
SOLUTION Using the -direction data, Equation 3.5a can be emploed to find the time t that the baseball is in the air: ( ) = t + a t = t since a = m/s Soling for t gies t = + 17. m = = +41. m/s.415 s 1 The displacement in the direction can now be ealuated b using the -direction data table and the alue of t =.415 s. Using Equation 3.5b, we hae ( ) ( ) ( ) ( ) = t + a t = + = 1 1 m/s.415 s 9.8 m/s.415 s.844 m The distance that the ball drops is gien b the magnitude of this result, so Distance =.844 m. WebAssign Problem 4: In the jaelin throw at a track-and-field eent, the jaelin is launched at a speed of 9 m/s at an angle of 36 aboe the horizontal. As the jaelin traels upward, its elocit points aboe the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 36 at launch to 18? REASONING As shown in the drawing, the angle that the elocit ector makes with the horizontal is gien b tan θ = θ where, from Equation 3.3b, = + a t = sin θ + a t and, from Equation 3.3a (since a = ), = = cos θ Therefore, SOLUTION Soling for t, we find sin θ + at tan θ = = cosθ
( cos θ tan θ sinθ ) ( 9 m/s ) ( cos 36 tan 18 sin 36 ) t = = = a 9.8 m/s.96 s WebAssign Problem 5: The lob in tennis is an effectie tactic when our opponent is near the net. It consists of lofting the ball oer his head, forcing him to moe quickl awa from the net (see the drawing). Suppose that ou loft the ball with an initial speed of 15. m/s, at an angle of 5. aboe the horizontal. At this instant our opponent is 1. m awa from the ball. He begins moing awa from ou.3 s later, hoping to reach the ball and hit it back at the moment that it is.1 m aboe its launch point. With what minimum aerage speed must he moe? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.) REASONING Using the data gien in the problem, we can find the maimum 1 flight time t of the ball using Equation 3.5b ( = t + a t ). Once the flight time is known, we can use the definition of aerage elocit to find the minimum speed required to coer the distance in that time. SOLUTION Equation 3.5b is quadratic in t and can be soled for t using the quadratic formula. According to Equation 3.5b, the maimum flight time is (with upward taken as the positie direction) 1 ( ) 1 ( ) ( ) ± ± + 4 a a t = = a a = ( ) ( ) 15. m/s sin 5. ± 15. m/s sin 5. +( 9.8 m/s ) (.1 m) 9.8 m/s =. s and.145 s
where the first root corresponds to the time required for the ball to reach a ertical displacement of = +.1 m as it traels upward, and the second root corresponds to the time required for the ball to hae a ertical displacement of = +.1 m as the ball traels upward and then downward. The desired flight time t is.145 s. During the.145 s, the horizontal distance traeled b the ball is [ ] = t = ( cos θ ) t = (15. m/s) cos 5. (.145 s) =.68 m Thus, the opponent must moe.68 m 1. m = 1.68 m in.145 s.3 s = 1.845 s. The opponent must, therefore, moe with a minimum aerage speed of min 1.68 m = = 1.845 s 5.79 m / s WebAssign Problem 6: Two cars, A and B, are traeling in the same direction, although car A is 186 m behind car B. The speed of A is 4.4 m/s, and the speed of B is 18.6 m/s. How much time does it take for A to catch B? REASONING Since car A is moing faster, it will eentuall catch up with car B. Each car is traeling at a constant elocit, so the time t it takes for A to catch up with B is equal to the displacement between the two cars ( = +186 m) diided b the elocit AB of A relatie to B. (If the relatie elocit were zero, A would neer catch up with B). We can find the elocit of A relatie to B b using the subscripting technique deeloped in Section 3.4 of the tet. AB = elocit of car A relatie to car B AG = elocit of car A relatie to the Ground = +4.4 m/s BG = elocit of car B relatie to the Ground = +18.6 m/s We hae chosen the positie direction for the displacement and elocities to be the direction in which the cars are moing. The elocities are related b AB = AG + GB SOLUTION The elocit of car A relatie to car B is AB = AG + GB = +4.4 m/s + ( 18.6 m/s) = +5.8 m/s, where we hae used the fact that GB = BG = 18.6 m/s. The time it takes for car A to catch car B is
+ 186 m t = = = 3.1 s +5.8 m/s AB WebAssign Problem 7: You are in a hot-air balloon that, relatie to the ground, has a elocit of 6. m/s in a direction due east. You see a hawk moing directl awa from the balloon in a direction due north. The speed of the hawk relatie to ou is. m/s. What are the magnitude and direction of the hawk s elocit relatie to the ground? Epress the directional angle relatie to due east. REASONING REASONING Let HB represent the elocit of the hawk relatie to the balloon and BG represent the elocit of the balloon relatie to the ground. Then, as indicated b Equation 3.7, the elocit of the hawk relatie to the ground is HG = HB + BG. Since the ectors HB and BG are at right angles to each other, the ector addition can be carried out using the Pthagorean theorem. SOLUTION Using the drawing at the right, we hae from the Pthagorean theorem, N O R T H = + HG HB BG = + = The angle θ is (. m / s) (6. m / s) 6.3 m / s F θ = tan H G I F KJ =. m / s H G I tan K J = 6. m / s 1 HB 1 BG 18, north of east H G θ B G H B E A S T WebAssign Problem 8: A person looking out the window of a stationar train notices that raindrops are falling erticall down at a speed of 5. m/s relatie to the ground. When the train moes at a constant elocit, the raindrops make an angle of 5 when the moe past the window, as the drawing shows. How fast is the train moing?
REASONING AND SOLUTION The elocit of the raindrops relatie to the train is gien b RT = RG + GT where RG is the elocit of the raindrops relatie to the ground and GT is the elocit of the ground relatie to the train. Since the train moes horizontall, and the rain falls erticall, the elocit ectors are related as shown in the figure at the right. Then R T 5 R G GT = RG tan θ = (5. m/s) (tan 5 ) =.3 m/s The train is moing at a speed of.3 m/s G T WebAssign Problem 9: Concept Questions (a) For a projectile that follows a trajector like that in Figure 3.1 the range is gien b R = t, where is the horizontal component of the launch elocit and t is the time of flight. Is proportional to the launch speed? (b) Is the time of flight t proportional to the launch speed? (c) Is the range R proportional to or? Eplain each answer. Problem In the absence of air resistance, a projectile is launched from and returns to ground leel. It follows a trajector similar to that in Figure 3.1 and has a range of 3 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle aboe the ground. What is the new range? CONCEPT QUESTIONS a. The horizontal component of the launch elocit is proportional to the launch speed, because = cos θ, where θ is the launch angle. b. The time of flight t is also proportional to the launch speed. The greater the launch speed, the longer the projectile is in flight. More eactl, we know that, for a projectile that is launched from and returns to ground leel, the ertical displacement is = m. Using Equation 3.5b, we hae 1 = t + a t 1 1 m = t + a t or m = + a t or t = a The flight time is proportional to the ertical component of the launch elocit, which, in turn, is proportional to the launch speed.
c. Since the range is gien b R = t and since both and t are proportional to, the range R is proportional to. SOLUTION The gien range is 3 m. When the launch speed doubles, the range increases b a factor of = 4, since the range is proportional to the square of the speed. Thus, the new range is Practice conceptual problems: b g R = b g = R = 4 3 m = 9 m 4 3 m 9 m 3. Is the acceleration of a projectile equal to zero when it reaches the top of its trajector? If not, wh not? REASONING AND SOLUTION As long as air resistance is negligible, the acceleration of a projectile is constant and equal to the acceleration due to grait. The acceleration of the projectile, therefore, is the same at eer point in its trajector, and can neer be zero. 5. A tennis ball is hit upward into the air and moes along an arc. Neglecting air resistance, where along the arc is the speed of the ball (a) a minimum and (b) a maimum? Justif our answers. REASONING AND SOLUTION The figure below shows the ball s trajector. The elocit of the ball (along with its and components) are indicated at three positions. As long as air resistance is neglected, we know that a = and a is the acceleration due to grait. Since a =, the component of the elocit remains the same and is gien b. The initial component of the elocit is and decreases as the ball approaches the highest point, where =. The magnitude of the component of the elocit then increases as the ball falls downward. Just before the ball strikes the ground, the elocit component is equal in magnitude to. The speed is the magnitude of the elocit. a. Since = when the ball is at the highest point in the trajector, the speed is a minimum there. b. Similarl, since is a maimum at the initial and final positions of the motion, the speed is a maimum at these positions.
θ θ f 8. A rifle, at a height H aboe the ground, fires a bullet parallel to the ground. At the same instant and at the same height, a second bullet is dropped from rest. In the absence of air resistance, which bullet strikes the ground first? Eplain. REASONING AND SOLUTION The two bullets differ onl in their horizontal motion. One bullet has =, while the other bullet has =. The time of flight, howeer, is determined onl b the ertical motion, and both bullets hae the same initial ertical elocit component ( = ). Both bullets, therefore, reach the ground at the same time. 13. On a rierboat cruise, a plastic bottle is accidentall dropped oerboard. A passenger on the boat estimates that the boat pulls ahead of the bottle b 5 meters each second. Is it possible to conclude that the boat is moing at 5 m/s with respect to the shore? Account for our answer. REASONING AND SOLUTION Since the plastic bottle moes with the current, the passenger is estimating the elocit of the boat relatie to the water. Therefore, the passenger cannot conclude that the boat is moing at 5 m/s with respect to the shore.